course MTH 163 Everything is clear on this assigment, except two points.Question 012 - •Quick response is clear, but I do not understand the MORE DETAILED RESPONSE (below). Can I assume this is covered in more detail in the supplemental link identified as Synopsis of the rate-of-change scheme of uniformly accelerated motion and graphical interpretation at Unif Accel Motion Scheme ?
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Given Solution: The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA Self-critique Rating:3 ********************************************* Question: `q003.If you make $60,000 per year then how much do you make per month? ********************************************* Your solution: $60,000/year = $60,000/12 months = $5,000/month Confidence Assessment: 3
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Given Solution: Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA Self-critique Rating: 3 ********************************************* Question: `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month? ********************************************* Your solution: An average of $5,000 per month would be correct. It is unlikely the business makes exactly 1/12 of the money for the year per month. There are frequently seasonal variations (income tax business would make more at filing time – construction business would make more during the summer, retail makes more from Thanksgiving through year end. Confidence Assessment: 3
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Given Solution: Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA Self-critique Rating:3 ********************************************* Question: `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate? ********************************************* Your solution: 300 miles/6 hours = 50 miles/1 hour or an average rate of 50 mph This does not mean that after two hours you would have traveled exactly 100 miles. The actual mph varies, (90 mph for an hour, then 10 mph after you get your speeding ticket, or more likely between 50 – 60 mph depending on road conditions (including the presence or absence of police cars). Actual speed could even be 80 mph, with stops of 0 mph factoring in to produce an average rate of travel. Confidence Assessment: 3
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Given Solution: The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA Self-critique Rating: 3 ********************************************* Question: `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled? ********************************************* Your solution: 60 gal / 1200 miles = .05 gallons per mile traveled as the question is worded. Usual format for this information is miles per gallon, which would be 1200 miles / 60 gal = 20 mpg Confidence Assessment: 3
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Given Solution: The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA Self-critique Rating: 3
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Given Solution: The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. Self-critique (if necessary) : NA Self-critique Rating: 3 ********************************************* Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup? ********************************************* Your solution: For each group, or for both groups combined? Per pushup or per ADDITIONAL pushup? The response would be different. As phrased it would be for both groups, so if I can assume the subgroups are of equal size (and therefore would receive equal weighting) 10 pushups + 50 pushups = 60 pushups for two groups Divide by 2, average pushups per day = 30 147 pounds + 162 pounds = 309 pounds increase in strength for both groups Divide by 2, average increase in lifting strength = 154.5 pounds 154.5 pounds / 30 pushups = 5.15 pounds average increase in lifting strength per daily pushup performed for a year. If I do not know (and cannot determine) that the test groups are of equal size and should receive equal weighting – no answer can be calculated for the group as a whole, and values must be calculated for each subgroup independently. Group 1: 147 pounds / 10 pushups = strength increase rate of 14.7 pounds / pushup This is much higher than the rate of return per pushup for the group as a whole. Group 2: 162 pounds / 50 pushups = strength increase rate of 3.24 pounds / pushup This would be a much lower return on effort invested, but greater overall increase in strength. If the question is actually the rate lifting strength increased per daily pushup over the base 10 daily pushups: (162-147) pounds / (50-10) = 15 pounds increase in lifting strength / 40 daily pushup increase This means the additional 40 daily pushups net an average increase in lifting strength of only .375 pounds each (Personally, I would lie down quietly until the urge to exercise had passed for that little return) Confidence Assessment: 3
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Given Solution: The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was not able to identify the outcome requested as the only interpretation of the question as worded, but I did complete the same calculation as one of the possible interpretations of this question. The question is restated below, with my additions in parentheses. “At the end of the year to(tal) lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup” (beyond the base rate of 10 per day)? Self-critique Rating:2 ********************************************* Question: `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight? ********************************************* Your solution: This question is phrased the same way as the last, so I will make the same adjustments as were implied in your last solution – restating the question as: “At what average rate did lifting strength increase (per pound) with respect to the added shoulder weight” (in excess of the base 10 pounds)? Therefore, 30 pound weight – 10 pound weight = 20 pound weight in excess of the base 10 pounds 188 pound increase in lifting weight-171 pound increase in lifting weight= 17 pound increase in lifting weight 17 pound increase in lifting weight / 20 pound weight in excess of the base 10 pounds = = 0.85 pound increase in lifting weight per pound of weight in excess of the base 10 pounds lifted daily Confidence Assessment: 2
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Given Solution: The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses. The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal. You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question. You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement. Self-critique Rating: 2 ********************************************* Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions? ********************************************* Your solution: First mark: (100 meters traveled / 12 seconds) Second mark: (200 meters traveled / 22 seconds) Interval between = Second mark – first mark = (200-100 meters traveled) / (22-12 seconds) = Average rate = 100 meters traveled / 10 seconds = 10 meters traveled/ second Confidence Assessment: 3
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Given Solution: The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Same Self-critique Rating: 3 ********************************************* Question: `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance? ********************************************* Your solution: Assuming (1) we are discussing the time required to cover the SECOND 100 meters distance: (2) speed at each mark was measured by radar gun at the point specified, rather than specifying an average rate over any given distance Given that the runner is moving at a decreasing rate, and assuming the rate of decrease was constant (unlikely, but all we have to go on), you would also assume an “average” speed of 10 meters/second and 9 meters / second, netting the average of 9.5 meters per second. Distance traveled = 100 meters at 9.5 meters per second = 100/9.5 seconds ≈ 10.5 seconds Confidence Assessment: 3
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Given Solution: At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): N/A Self-critique Rating: 3 ********************************************* Question: `q012. We just averaged two quantities, adding them and dividing by 2, to find an average rate. We didn't do that before. Why we do it now? ********************************************* Your solution: If given a total (distance traveled, time spent, etc) there is nothing to add. If given multiple rate (mpg, mph, etc) you must add them and divide by the number of entries to obtain an average. Confidence Assessment: 3
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Given Solution: In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. STUDENT QUESTION: I thought the change of an accumulating quantity was the rate? INSTRUCTOR RESPONSE: Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero). More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B. For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): • Quick response is clear, but I do not understand the MORE DETAILED RESPONSE (below). Can I assume this is covered in more detail in the supplemental link identified as Synopsis of the rate-of-change scheme of uniformly accelerated motion and graphical interpretation at Unif Accel Motion Scheme More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B. For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x. Self-critique Rating: 2